It's been a while since I've been in school (10yrs). I'm a little rusty at all my math. I wonder if someone could prove how the first formula is the same as the second.
Cheers!!!
$\displaystyle Vf^2 = Vi^2 +2a\Delta t $
Subtract $\displaystyle Vi^2$ from both sides
$\displaystyle Vf^2 - Vi^2 = 2a\Delta t $
Divide both sides by $\displaystyle 2\Delta t$
$\displaystyle a = \frac{Vf^2 - Vi^2}{2\Delta t}$
I see that this is not the same as the second formula you posted. Yours was $\displaystyle a = \frac{Vf^2 - Vi^2}{2\boxed{a}\Delta t}$
I don't even have an idea what formula is this. It looks like the change of position formula of a constant acceleration, without time. Which is,
$\displaystyle \Delta x$ as the change of position,
$\displaystyle V_1$ and $\displaystyle V_2$ as the first and last velocities,
$\displaystyle a$ as the constant acceleration
$\displaystyle \Delta x = \frac{V_2^2-V_1^2}{2a}$
Your absolutely right there was a typo in the second equation. I was quite tired when I entered the equation. The equation involves distance rather than " t " time as well. My second lesson here is the importance of proof reading my equations. Thanks again.
Cheers!!!